Let me add the details for Henri Cohen's nice answer, without claiming any originality. We have $$\sum_{m,n\geq 1}\frac{\cos\left(\frac{m}n\right)}{m^2n^2}=\zeta(4)\sum_{(a,b)=1}\frac{\cos\left(\frac{a}b\right)}{a^2b^2}.$$ On the other hand, by the identity (cf. Ron Gordon's answer here) $$\sum_{m\geq 1}\frac{\cos(mx)}{m^2}=\frac{x^2}{4}-\frac{\pi x}{2}+\zeta(2),\qquad 0\leq x\leq 2\pi,$$ we also have $$\sum_{m,n\geq 1}\frac{\cos\left(\frac{m}n\right)}{m^2n^2}=\sum_{n\geq 1}\frac{1}{n^2}\left(\frac{n^{-2}}{4}-\frac{\pi n^{-1}}{2}+\zeta(2)\right)=\frac{\zeta(4)}{4}-\frac{\pi\zeta(3)}{2}+\zeta(2)^2.$$ Comparing the left hand sides of the first and third equation, we conclude that $$\sum_{(a,b)=1}\frac{\cos\left(\frac{a}b\right)}{a^2b^2}=\frac{1}{4}-\frac{\pi\zeta(3)}{2\zeta(4)}+\frac{\zeta(2)^2}{\zeta(4)}.$$ Here we have $\zeta(2)=\pi^2/6$ and $\zeta(4)=\pi^4/90$, therefore $$\sum_{(a,b)=1}\frac{\cos\left(\frac{a}b\right)}{a^2b^2}=\frac{11}{4}-\frac{45\zeta(3)}{\pi^3}.$$

Let me add the details for Henri Cohen's nice answer, without claiming any originality. We have $$\sum_{m,n\geq 1}\frac{\cos\left(\frac{m}n\right)}{m^2n^2}=\zeta(4)\sum_{(a,b)=1}\frac{\cos\left(\frac{a}b\right)}{a^2b^2}.$$ On the other hand, by the identity (cf. Ron Gordon's answer here) $$\sum_{m\geq 1}\frac{\cos(mx)}{m^2}=\frac{x^2}{4}-\frac{\pi x}{2}+\zeta(2),\qquad 0\leq x\leq 2\pi,$$ we also have $$\sum_{m,n\geq 1}\frac{\cos\left(\frac{m}n\right)}{m^2n^2}=\sum_{n\geq 1}\frac{1}{n^2}\left(\frac{n^{-2}}{4}-\frac{\pi n^{-1}}{2}+\zeta(2)\right)=\frac{\zeta(4)}{4}-\frac{\pi\zeta(3)}{2}+\zeta(2)^2.$$ Comparing the right hand sides of the first and third equation, we conclude that $$\sum_{(a,b)=1}\frac{\cos\left(\frac{a}b\right)}{a^2b^2}=\frac{1}{4}-\frac{\pi\zeta(3)}{2\zeta(4)}+\frac{\zeta(2)^2}{\zeta(4)}.$$ Here we have $\zeta(2)=\pi^2/6$ and $\zeta(4)=\pi^4/90$, therefore $$\sum_{(a,b)=1}\frac{\cos\left(\frac{a}b\right)}{a^2b^2}=\frac{11}{4}-\frac{45\zeta(3)}{\pi^3}.$$

Let me add the details for Henri Cohen's nice answer, without claiming any originality. We have $$\sum_{m,n\geq 1}\frac{\cos\left(\frac{m}n\right)}{m^2n^2}=\zeta(4)\sum_{(a,b)=1}\frac{\cos\left(\frac{a}b\right)}{a^2b^2}.$$ On the other hand, by the identity $$\sum_{m\geq 1}\frac{\cos(mx)}{m^2}=\frac{x^2}{4}-\frac{\pi x}{2}+\zeta(2),\qquad 0\leq x\leq 2\pi,$$ we also have $$\sum_{m,n\geq 1}\frac{\cos\left(\frac{m}n\right)}{m^2n^2}=\sum_{n\geq 1}\frac{1}{n^2}\left(\frac{n^{-2}}{4}-\frac{\pi n^{-1}}{2}+\zeta(2)\right)=\frac{\zeta(4)}{4}-\frac{\pi\zeta(3)}{2}+\zeta(2)^2.$$ Comparing the left hand sides of the first and third equation, we conclude that $$\sum_{(a,b)=1}\frac{\cos\left(\frac{a}b\right)}{a^2b^2}=\frac{1}{4}-\frac{\pi\zeta(3)}{2\zeta(4)}+\frac{\zeta(2)^2}{\zeta(4)}.$$ Here we have $\zeta(2)=\pi^2/6$ and $\zeta(4)=\pi^4/90$, therefore $$\sum_{(a,b)=1}\frac{\cos\left(\frac{a}b\right)}{a^2b^2}=\frac{11}{4}-\frac{45\zeta(3)}{\pi^3}.$$

Let me add the details for Henri Cohen's nice answer, without claiming any originality. We have $$\sum_{m,n\geq 1}\frac{\cos\left(\frac{m}n\right)}{m^2n^2}=\zeta(4)\sum_{(a,b)=1}\frac{\cos\left(\frac{a}b\right)}{a^2b^2}.$$ On the other hand, by the identity (cf. Ron Gordon's answer here) $$\sum_{m\geq 1}\frac{\cos(mx)}{m^2}=\frac{x^2}{4}-\frac{\pi x}{2}+\zeta(2),\qquad 0\leq x\leq 2\pi,$$ we also have $$\sum_{m,n\geq 1}\frac{\cos\left(\frac{m}n\right)}{m^2n^2}=\sum_{n\geq 1}\frac{1}{n^2}\left(\frac{n^{-2}}{4}-\frac{\pi n^{-1}}{2}+\zeta(2)\right)=\frac{\zeta(4)}{4}-\frac{\pi\zeta(3)}{2}+\zeta(2)^2.$$ Comparing the left hand sides of the first and third equation, we conclude that $$\sum_{(a,b)=1}\frac{\cos\left(\frac{a}b\right)}{a^2b^2}=\frac{1}{4}-\frac{\pi\zeta(3)}{2\zeta(4)}+\frac{\zeta(2)^2}{\zeta(4)}.$$ Here we have $\zeta(2)=\pi^2/6$ and $\zeta(4)=\pi^4/90$, therefore $$\sum_{(a,b)=1}\frac{\cos\left(\frac{a}b\right)}{a^2b^2}=\frac{11}{4}-\frac{45\zeta(3)}{\pi^3}.$$